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The partition function

$$Z_N = \left( \prod_{n=2}^N \int_{-\infty}^{+\infty} \mathrm{d} x_n \right) \exp \left[ - \frac{\beta K}{2} \sum_{i=1}^{N-1} (x_{i+1} + x_i)^2 \right]$$

can be re-expressed as

\begin{eqnarray*}
Z_N = \int_{-\infty}^{+\infty} \mathrm{d} x_2 \, e^{- \beta K x_2^2 / 2}
\int_{-\infty}^{+\infty} \mathrm{d} x_3 \, e^{- \beta K (x_3 - x_2)^2 / 2} ...\\
... \int_{-\infty}^{+\infty} \mathrm{d} x_{N-1} \, e^{- \beta K (x_{N-1} - x_{N-2})^2 / 2}
\int_{-\infty}^{+\infty} \mathrm{d} x_N \, e^{- \beta K (x_N - x_{N-1})^2 / 2}
\end{eqnarray*}

i.e. a product of gaussian integrals. Each integral can be evaluated directly via the substitution $\xi_i = x_i - x_{i-1}$

$$\int_{-\infty}^{+\infty} \mathrm{d} \xi \, e^{-\beta K \xi^2 / 2} = \sqrt{ \frac{2\pi}{\beta K} }$$

there are a total of $N-1$ such integrals, for a partition function

$$Z_N = \left( \frac{2\pi}{\beta K} \right)^{\frac{N-1}{2}}$$

The average energy

$$\langle E \rangle = - \frac{ \partial }{ \partial \beta } \log Z_N = \frac{N-1}{2} \frac{ \partial }{ \partial \beta } \log \beta = \frac{N-1}{2} k_B T$$

The equipartition theorem states that the average energy of a system with $M$ degrees of freedom is $M k_B T / 2$. For the gaussian chain there are $N$ masses, but the first mass is fixed, hence the chain has $N-1$ degrees of freedom, and its average energy is in agreement with the equipartition theorem.

The probability that the $N$th bead is at position $y$ is given (schematically) by

$$\mathbb{P}_N(y) = \sum_{\mathrm{conf}} \delta(y-x_N) \frac{ e^{-\beta \mathcal{H}} }{Z_N}$$

referring to the expression for $Z_N$, and using the integral representation of the $\delta$-function i.e.

$$\delta(y - x_n) = \frac{1}{2 \pi} \int \mathrm{d} k \, \exp \left\{ ik \left( y - \sum_i^{N-1} \xi_i \right) \right\}$$

\begin{eqnarray*}
\mathbb{P}_N(y) &=& \frac{1}{2 \pi Z_N} \int \mathrm{d} k \,e^{iky} \int \mathrm{d} \xi_1 \, e^{- \beta K \xi_1^2 / 2} \int \mathrm{d} \xi_2 \, e^{- \beta K \xi_2^2 / 2} \cdots \\
&& \cdots \int \mathrm{d} \xi_{N-1} \, e^{- \beta K \xi_{N-1}^2 / 2} e^{ik\xi_1} e^{ik\xi_2} \cdots e^{ik\xi_{N-1}} \\
&=& \frac{1}{2 \pi Z_N} \int \mathrm{d} k \, e^{iky} \left[ \int \mathrm{d} \xi \, \exp\{- \beta K \xi^2 / 2 + ik \xi \} \right]^{N-1} \\
&=& \frac{1}{2 \pi Z_N} \int \mathrm{d} k \, e^{iky} \left[ \sqrt{ \frac{2 \pi}{\beta K} } \exp \left\{ \frac{ik}{2 \beta K} \right\} \right]^{N-1} \\
&=& \frac{1}{2 \pi} \int \mathrm{d} k \, \exp \left\{ - \frac{N-1}{2 \beta K} k^2 + iky \right\} \\
&=& \sqrt{ \frac{2 \beta K \pi}{2 (N-1) \pi} } \exp \left\{ \frac{ - \beta K }{ 2(N-1) } y^2 \right\}
\end{eqnarray*}

So the probability distribution for the total chain length is a normalised gaussian, with variance proportional to the number of springs and inversely proportional to the stiffness $K$ of the individual springs, as we would expect.

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